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The reaction between aluminium and iron(iii) oxide is represented by the equation below 2Al + Fe2O3 -Al2O3 +2Fe in one process 124g of All are reacted with 601g of Fe2O3. a. Calculate the mass of Al2O3 formed b. How much of the excess reagent is left at the end of the reaction
one year ago

## Answers : (1)

Shweta Ahire
44 Points

124gm of Al means 124/27 moles=4.6 moles
601gm of fe2O3= 601/160 moles = 3.75 moles
So from this we can say that Al is limiting reagent
Hence all Al will be consume and mole of Al2O3 formed is half of the mole of Al
So moles of Al2O3 formed is= 1/2×4.6
= 2.3 moles
Mass of  Al2O3 formed   = 2.3 × 102= 234.6 gm

Fe2O3 is excess reagent so some part of it consumes and some part left
Mole consumed of Fe2O3 is =1/2 × mole of Al
=1/2× 4.6 = 2.3
So remaining moles = 3.75 - 2.3 = 1.45
Mass = 1.45× 160 = 232gm
So 232gm of Fe2O3 (excess reagent) is left at end of reaction.

one year ago
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